A280050 a(n) = Sum_{k=2..n} k/lpf(k), where lpf(k) is the least prime dividing k (A020639).
0, 1, 2, 4, 5, 8, 9, 13, 16, 21, 22, 28, 29, 36, 41, 49, 50, 59, 60, 70, 77, 88, 89, 101, 106, 119, 128, 142, 143, 158, 159, 175, 186, 203, 210, 228, 229, 248, 261, 281, 282, 303, 304, 326, 341, 364, 365, 389, 396, 421, 438, 464, 465, 492, 503, 531, 550, 579, 580, 610, 611, 642, 663, 695, 708, 741, 742, 776, 799, 834, 835
Offset: 1
Examples
For n = 8 the divisors of the first eight positive integers are {1}, {1, 2}, {1, 3}, {1, 2, 4}, {1, 5}, {1, 2, 3, 6}, {1, 7}, {1, 2, 4, 8}, so a(8) = 1 + 1 + 2 + 1 + 3 + 1 + 4 = 13.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Least Prime Factor.
- Eric Weisstein's World of Mathematics, Proper Divisor.
Programs
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Mathematica
Table[Sum[k/FactorInteger[k][[1, 1]], {k, 2, n}], {n, 71}] Join[{0}, Accumulate[Table[k/FactorInteger[k][[1, 1]], {k, 2, 71}]]] (* Amiram Eldar, Jul 03 2025 *)
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PARI
list(kmax) = {my(s = 0); print1(s, ", "); for(k = 2, kmax, s += k/factor(k)[1,1]; print1(s, ", "));} \\ Amiram Eldar, Jul 03 2025
Formula
a(n) = Sum_{k=2..n} k/A020639(k).
a(n) + 1 = Sum_{k=1..n} A032742(k).
a(p^k) = a(p^k-1) + p^(k-1), when p is prime.
a(n) ~ c * n^2, where c = (1/2) * Sum_{k>=1} A005867(k-1)/(prime(k)*A002110(k)) = 0.165049... . - Amiram Eldar, Jul 03 2025
Comments